Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I read that one can prove by AM-GM-inequality that for all $a,b,c\in\mathbb{R}_+$ we have that $$11(a^6 + b^6 + c^6) + 40abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$ How this can be done? Is it possible to prove stronger inequalities like $$10(a^6 + b^6 + c^6) + 41abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a),$$ or even $$a^6 + b^6 + c^6 + 50abc(ab^2 + bc^2 + ca^2) \ge 51abc(a^2b + b^2c + c^2a)$$ without computer?

share|cite|improve this question
The basic idea is that one can apply (weighted) AM-GM to take convex combinations of exponent vectors. This technique is also called "bunching," or Muirhead's inequality:'s_inequality – Qiaochu Yuan Nov 21 '10 at 12:46
up vote 2 down vote accepted

You want to apply AM-GM to some of the terms on the LHS to get some of the terms on the RHS. A naive way to do this such as

$$b^6 + c^6 + 2a^2 b^3 c \ge 4 a b^3 c^2$$

uses too many of the $11$ terms relative to the $40$ terms and only proves the weaker inequality

$$a^6 + b^6 + c^6 + abc(ab^2 + bc^2 + ca^2) \ge 2abc(a^2 b + b^2 c + c^2 a)$$

after cyclic summation. A slightly better combination of terms is

$$5a^6 + c^6 + 12a^2 b^3 c \ge 18 a^3 b^2 c$$

but this still uses too many of the $11$ terms relative to the $40$ terms and proves the inequality

$$a^6 + b^6 + c^6 + 2abc(ab^2 + bc^2 + ca^2) \ge 3abc(a^2 b + b^2 c + c^2 a)$$

after cyclic summation. You get a strong enough result by mixing some of the $40$ terms with each other to get

$$4a^6 + 12a^2 b^3 c + 4a^3 b c^2 \ge 20 a^3 b^2 c$$

which proves

$$a^6 + b^6 + c^6 + 4abc(ab^2 + bc^2 + ca^2) \ge 5abc(a^2 b + b^2 c + c^2 a).$$

Now multiply by $10$ and apply AM-GM to the remaining terms to conclude. The question of whether one can do better than this using AM-GM is a geometric question about using convex linear combinations of the vectors $(6, 0, 0), (0, 6, 0), (0, 0, 6), (2, 3, 1), (3, 1, 2), (1, 2, 3)$ to obtain the vectors $(3, 2, 1), (2, 1, 3), (1, 3, 2)$ and one might be able to prove that the above result is optimal.

share|cite|improve this answer

I don't have enough karma to comment, so I'll just drop this here: discusses the problem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.